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Question Number 38454 by maxmathsup by imad last updated on 25/Jun/18

let f(x)=∫_0 ^∞ ((1−cos(xt^2 ))/t^2 ) e^(−xt^2 ) dt with x>0 1) find a simple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(2t^2 ))/t^2 ) e^(−3t^2 ) dt .

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−{xt}^{\mathrm{2}} } {dt}\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−\mathrm{3}{t}^{\mathrm{2}} } {dt}\:. \\ $$

Commented bymath khazana by abdo last updated on 26/Jun/18

the Q is f(x)=∫_0 ^∞ ((1−cos(at^2 ))/t^2 ) e^(−xt^2 ) dt

$${the}\:{Q}\:{is}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({at}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−{xt}^{\mathrm{2}} } {dt} \\ $$

Commented byabdo.msup.com last updated on 27/Jun/18

wehave f(x)=∫_0 ^∞ ((1−cos(at^2 ))/t^2 ) e^(−xt^2 ) dt⇒ f^′ (x)= −∫_0 ^∞ (1−cos(at^2 )e^(−xt^2 ) dt =∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt −∫_0 ^∞ e^(−xt^2 ) dt but ∫_0 ^∞ e^(−xt^2 ) dt =_((√x)t=u) ∫_0 ^∞ e^(−u^2 ) (du/(√x)) =(1/(√x))∫_0 ^∞ e^(−u^2 ) du =(π/(2(√x))) and ∫_0 ^∞ cos(at^2 )e^(−xt^2 ) dt=(1/2) ∫_(−∞) ^(+∞) cos(at^2 )e^(−xt^2 ) dt =(1/2) Re( ∫_(−∞) ^∞ e^(iat^2 −xt^2 ) dt) but ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt =_((√(−x+ia))t=u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/(√(−x+ia))) =((√π)/(√(−x+ia))) but −x+ia=(√(x^2 +a^2 )){((−x)/(√(x^2 +a^2 ))) +((ia)/(√(x^2 +a^2 )))} =r e^(iθ) ⇒r=(√(x^2 +a^2 )) and tanθ=−(a/x) ⇒ θ=−arctan((a/x))⇒−x+ia=r e^(−iarctan((a/x))) ⇒(√(−x+ia))=(x^2 +a^2 )^(1/4) e^(−(i/2)arctan((a/x))) ⇒ ∫_(−∞) ^(+∞) e^((−x+ia)t^2 ) dt=(√π) (x^2 +a^2 )^(−(1/4)) e^((i/2)arctan((a/x))) f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/x)))⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt +c

$${wehave}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{cos}\left({at}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−{xt}^{\mathrm{2}} } {dt}\Rightarrow \\ $$ $${f}^{'} \left({x}\right)=\:−\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}−{cos}\left({at}^{\mathrm{2}} \right){e}^{−{xt}^{\mathrm{2}} } {dt}\right. \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \:\:{cos}\left({at}^{\mathrm{2}} \right){e}^{−{xt}^{\mathrm{2}} } {dt}\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{xt}^{\mathrm{2}} } {dt}\:{but} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}^{\mathrm{2}} } {dt}\:=_{\sqrt{{x}}{t}={u}} \int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\sqrt{{x}}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{{x}}}\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } {du}\:=\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\:{and} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:{cos}\left({at}^{\mathrm{2}} \right){e}^{−{xt}^{\mathrm{2}} } {dt}=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:{cos}\left({at}^{\mathrm{2}} \right){e}^{−{xt}^{\mathrm{2}} } {dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:{Re}\left(\:\int_{−\infty} ^{\infty} \:\:{e}^{{iat}^{\mathrm{2}} −{xt}^{\mathrm{2}} } {dt}\right)\:{but} \\ $$ $$\int_{−\infty} ^{+\infty} \:\:{e}^{\left(−{x}+{ia}\right){t}^{\mathrm{2}} } {dt}\:=_{\sqrt{−{x}+{ia}}{t}={u}} \int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\sqrt{−{x}+{ia}}} \\ $$ $$=\frac{\sqrt{\pi}}{\sqrt{−{x}+{ia}}}\:\:{but}\: \\ $$ $$−{x}+{ia}=\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\left\{\frac{−{x}}{\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }}\:+\frac{{ia}}{\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }}\right\} \\ $$ $$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\:{and}\:{tan}\theta=−\frac{{a}}{{x}}\:\Rightarrow \\ $$ $$\theta=−{arctan}\left(\frac{{a}}{{x}}\right)\Rightarrow−{x}+{ia}={r}\:{e}^{−{iarctan}\left(\frac{{a}}{{x}}\right)} \\ $$ $$\Rightarrow\sqrt{−{x}+{ia}}=\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{x}}\right)} \Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\:{e}^{\left(−{x}+{ia}\right){t}^{\mathrm{2}} } {dt}=\sqrt{\pi}\:\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{x}}\right)} \\ $$ $${f}^{'} \left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{x}}\right)\right)\Rightarrow \\ $$ $${f}\left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{.} ^{{x}} \left({t}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} {cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{t}}\right)\right){dt} \\ $$ $$+{c} \\ $$ $$ \\ $$

Commented byabdo.msup.com last updated on 27/Jun/18

error at the final lines f^′ (x)=((√π)/2)(x^2 +a^2 )^(−(1/4)) cos{(1/2)arctan((a/x))} −(π/(2(√x))) ⇒ f(x)=((√π)/2) ∫_. ^x (t^2 +a^2 )^(−(1/4)) cos((1/2)arctan((a/t)))dt −π(√x) +c

$${error}\:{at}\:{the}\:{final}\:{lines} \\ $$ $${f}^{'} \left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left\{\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{x}}\right)\right\} \\ $$ $$−\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow \\ $$ $${f}\left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{.} ^{{x}} \left({t}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} {cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{a}}{{t}}\right)\right){dt} \\ $$ $$−\pi\sqrt{{x}}\:\:\:+{c} \\ $$

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